Sorry Mariano, I was trying to use the fact that $m$ is the slope of the line and try to somehow compare it with the derivative of the curve at those three points of intersections, i.e. $x_1, x_2, x_3$ I don't think this is a good approach as I have been thinking about it for some times. Do you have any other approach in mind?
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bonyankanFeb 13 '12 at 7:16

Do you know what the characteristic of a field is?
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Will JagyFeb 13 '12 at 7:25

1 Answer
1

I will work here over a field of characteristic zero, and in fact, to fix ideas, I will be working just over $\mathbb{Q}$ (you can work out what may happen if the characteristic of the field of definition is positive and $>3$... not much changes). I will also only work out here the following case: if a line and an elliptic curve intersect in two points, the line is tangent to the curve.

Let $y=f(x)$ be a function, differentiable at a point $x=a$. Let us define tangent as follows: we say that a line $L: y=g(x)=mx+n$ is tangent to the graph of $f(x)$ at $x=a$, if
$$g(a)=f(a)\quad \text{ and }\quad g'(a)=f'(a).$$

With this definition in mind, let $E/\mathbb{Q}$ be an elliptic curve given by $y^2=x^3+Ax+B$, let $y=f(x)=\sqrt{x^3+Ax+B}$ and let $L: y=g(x)=mx+n$ be a line such that the intersection of $L$ and $E$ intersect exactly at 2 points, $(a_1,b_1)$ and $(a_2,b_2)$. This means that
$$ g(x)^2 - f(x)^2$$
is a polynomial of degree $3$ with only $2$ roots, namely $a_1$ and $a_2$, so one of them is a repeated root. Let us say $a_1$ is repeated, then
$$ g(x)^2 - f(x)^2=(x-a_1)^2(x-a_2).$$
Thus, if we differentiate at a point $x$ we obtain:
$$ 2g(x)g'(x) - 2f(x)f'(x)=(x-a_1)(2(x-a_2)+(x-a_1)). \quad (\diamond)$$

If $b_1\neq 0$, evaluate ($\diamond$) at $x=a_1$. Note that $f(a_1)=g(a_1)=b_1$ by assumption, because $E$ and $L$ intersect at $(a_1,b_1)$. If $b_1\neq 0$, then $g'(a_1)=f'(a_1)$, and therefore $L$ is tangent to the graph of $f(x)$ at $x=a_1$ or, equivalently, $L$ is tangent to $E$ at $(a_1,b_1)$.

Otherwise, if $b_1=0$, then $f(x)$ is not differentiable at $x=a_1$, and in fact either (i) $\lim_{x\to a_1^+} f'(x)=\infty$ and $f(x)$ is not defined on $(a_1,\epsilon)$, or (ii) $\lim_{x\to a_1^-} f'(x)=\infty$ and $f(x)$ is not defined on $(\epsilon,a_1)$, for some $\epsilon>0$. Either way, by taking an appropriate limit in ($\diamond$) we find that $\lim_{x\to a_1} g'(x) = \infty$ which implies that $L$ must be a vertical line $x=a_1$. Since in this case ($b_1=0$) the tangent line to $E$ is also $x=a_1$, we conclude that $L$ is tangent to $E$, as claimed.